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Cauchy sequence proof.

by cragar
Tags: cauchy, proof, sequence
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cragar
#1
Mar4-12, 12:57 AM
P: 2,465
1. The problem statement, all variables and given/known data
Assume [itex] x_n [/itex] and [itex] y_n [/itex] are Cauchy sequences.
Give a direct argument that [itex] x_n+y_n [/itex] is Cauchy.
That does not use the Cauchy criterion or the algebraic limit theorem.
A sequence is Cauchy if for every [itex] \epsilon>0 [/itex] there exists an
[itex] N\in \mathbb{N} [/itex] such that whenever [itex] m,n\geq N [/itex]
it follows that [itex] |a_n-a_m|< \epsilon [/itex]
3. The attempt at a solution
Lets call [itex] x_n+y_n=c_n [/itex]
now we want to show that [itex] |c_m-c_n|< \epsilon [/itex]
Lets assume for the sake of contradiction that
[itex] c_m-c_n> \epsilon [/itex]
so we would have
[itex]|x_m+y_m-x_n-y_n|> \epsilon [/itex]
[itex] x_m> \epsilon+y_n-y_m [/itex]
since [itex] y_n>y_m [/itex]
and we know that [itex] x_m< \epsilon [/itex]
so this is a contradiction and the original statement must be true.
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Office_Shredder
#2
Mar4-12, 01:08 AM
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Quote Quote by cragar View Post
1. The problem statement, all variables and given/known data
Assume [itex] x_n [/itex] and [itex] y_n [/itex] are Cauchy sequences.
Give a direct argument that [itex] x_n+y_n [/itex] is Cauchy.
That does not use the Cauchy criterion or the algebraic limit theorem.
A sequence is Cauchy if for every [itex] \epsilon>0 [/itex] there exists an
[itex] N\in \mathbb{N} [/itex] such that whenever [itex] m,n\geq N [/itex]
it follows that [itex] |a_n-a_m|< \epsilon [/itex]
3. The attempt at a solution
Lets call [itex] x_n+y_n=c_n [/itex]
now we want to show that [itex] |c_m-c_n|< \epsilon [/itex]
It's always good to be careful with wording. We want to show that if n and m are big enough, that this inequality holds.

since [itex] y_n>y_m [/itex]
This is probably not true, especially since you haven't even said what n and m are besides arbitrary numbers!

and we know that [itex] x_m< \epsilon [/itex]
This is also probably not true since there's no reason to think the limit is zero

To get the contradiction you're going to want to use the triangle inequality on |(xn-xm)+(yn-ym)|
cragar
#3
Mar4-12, 03:48 PM
P: 2,465
ok thanks for your response.
So I take [itex] |(x_n-x_m)+(y_n-y_m)| \leq |x_n-x_m|+|y_n-y_m| [/itex]
lets assume that [itex] |x_n-x_m|+|y_n-y_m| > \epsilon [/itex]
Im going to rewrite it as [itex] A+B> \epsilon [/itex]
so now we have [itex] A> \epsilon -B [/itex]
Can I just say this since we know that [itex] A< \epsilon [/itex] and [itex]B< \epsilon [/itex] since [itex] \epsilon [/itex] can be any number bigger than zero, then both of these values should be less than [itex] \frac{\epsilon}{2} [/itex]
therefore [itex] A+B< \epsilon [/itex]
I have a feeling my last step is not ok

Office_Shredder
#4
Mar4-12, 03:55 PM
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Cauchy sequence proof.

Quote Quote by cragar View Post
Can I just say this since we know that [itex] A< \epsilon [/itex] and [itex]B< \epsilon [/itex] since [itex] \epsilon [/itex] can be any number bigger than zero, then both of these values should be less than [itex] \frac{\epsilon}{2} [/itex]
therefore [itex] A+B< \epsilon [/itex]
This is the crux of the argument. It's not the whole proof of course - A and B aren't always that small. Feel free to post a full proof if you want it checked over for errors
cragar
#5
Mar4-12, 03:58 PM
P: 2,465
Ok , Am I thinking about this in the right way.
Office_Shredder
#6
Mar4-12, 04:10 PM
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Well, I can't read your mind but the part I quoted is the basis of the a correct argument for the proof. You just have to add the fact that these are sequences - A and B aren't always that small, but as long as n and m are big enough they are (by definition of the x's and the y's forming Cauchy sequences)


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